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I have this question and it seems like I am stuck somewhere. Could someone help me with this? It involves parametric curves with the parameter of the curve in terms of $\pi$.

Find the cartesian equation of the curve, sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases. [Verify using Mathematica]
i). $\;\;\;x = \sin \frac{1}{2}\theta, \;\; y = \cos \frac{1}{2}\theta, \;\; -\pi \leq \theta \leq \pi.$
ii). $\;\;x = \frac{1}{2}\cos \theta, \;\; y = 2 \sin \theta, \;\; 0 \leq \theta \leq \pi.$

Dave L. Renfro
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Eron Kumar
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  • Effort? Don’t just post your homework with zero effort. – Ted Shifrin Mar 29 '24 at 03:35
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    Seconding @TedShifrin remark - it helps if you show what you have managed to do already and ask a more specific question. – Red Five Mar 29 '24 at 03:43
  • Look up parameterization of conic sections. A hit. The area of the figures described have Area=1. – TurlocTheRed Mar 29 '24 at 04:19
  • iam sorry @TedShifrin – Eron Kumar Mar 29 '24 at 08:16
  • thank you @TurlocTheRed ill look that up – Eron Kumar Mar 29 '24 at 08:18
  • @RedFive thank you so much for the approach. this is an assignment question however i have been looking over my lecture notes and couldnt find similar examples to work with. i tried to search on youtube too but they seem to be just general soltuions – Eron Kumar Mar 29 '24 at 08:21
  • Although this may be a problem in your calculus class, it's really a precalculus/parametric-curve topic that involves no calculus at all. Also, you called this "Advanced Calculus II", which might for some reason be the title of your course (the problem itself is, however, a standard problem found in a regular (not advanced, not honors) U.S. calculus 2 course), but for many people here the use of "Advanced Calculus" would be very misleading (because the name usually refers to a 2-semester U.S. upper level undergraduate course). – Dave L. Renfro Mar 29 '24 at 08:49

1 Answers1

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Here is one approach for part (i):

$x^{2}=\sin^{2}{\frac{\theta}{2}}$, $y^{2}=\cos^{2}{\frac{\theta}{2}}$

$x^{2}+y^{2}=\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}}=1$

When $\theta=-\pi$, $x=\sin{\frac{-\pi}{2}}=-1$ and when $\theta=\pi$, $x=\sin{\frac{\pi}{2}}=1$ so the domain is $[-1,1]$

For part (ii), we need $\sin{x}$ and $\cos{x}$ to both have the same coefficient, so start by writing

$2x=\cos{\theta}$ and $\frac{y}{2}=\sin{\theta}$
Then $(2x)^{2}+(\frac{y}{2})^{2}=\sin^{2}{\theta}+\cos^{2}{\theta}=1$
So $4x^{2}+\frac{y^{2}}{4}=1$

The domain can be determined from the equation for $x$ in a similar way to part (i).

This is only one possible approach. There are others.

I will leave the direction of the curve and the verification using Mathematica up to you, since this looks a lot like homework...

(Hint though for Mathematica, the Manipulate command can be very useful here...)

Red Five
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